sprotte said:
Hi all,
after 10 years without any physics, I have trouble solving the following problem:
"If your mean surface temperature is 28 C, what is your emittance? If the mean wall temperature in the room in which you are standing is 20 C, what is the average thermal irradiance at your surface? Estimate your net radiant heat loss. Assume surface emissivity is 0.97 and the wall emissivity is 1."I know that I have to calculate the emittance with the Stephan-Boltzmann-Law, but I have no clue how to get the irradiance at the surface and how to estimate the net radiant heat loss! Should not be that much of a problem for a physics expert, but I am struggling with this for hours now. Thanks in advance...
This thread was created over 12 years ago. The reason it took so long is because the OP did not show work. After "hours" of struggle, there should be at least a little work to show. For present and future readers who want help: if you're stuck, show us what you've tried and where you're stuck. Only then can we help.
But since this thread is so old, and the OP has surely moved on, I'll go ahead and give the steps to solve this sort of problem:
Radiant emittance, M_e, is the power per unit area radiated by a body (example units are \mathrm{W/m^2}). For a given type of surface, having a known emissivity, it is dependent only on temperature.
Emissivity, \varepsilon, varies between 0 and 1 such that \varepsilon = 1 for a perfectly black body and \varepsilon = 0 for an idealized [albeit unrealistic] perfectly white surface that doesn't radiate at all.
Radiated emittance, M_e can be calculated using the formula (based on the Stefan-Boltzmann law, modified for non-black-body surfaces having an emissivity):
M_e = \varepsilon \sigma T^4,
where \sigma is the Stefan-Boltzmann constant and T is the absolute temperature
So for your emittance, you have all the information you need. The variable \varepsilon = 0.97, and the temperature is 28 deg C + 273.15 K = 301.15 K.
The Stefan-Boltzmann constant is \sigma = 5.670374 \times 10^{-8} \ \mathrm{[W \cdot m^{-2} \cdot K^{-4}]}
That's enough for your own radiant emittance.
As for the room, you need to assume that you are surrounded by walls that all have the same emissivity of 1, including the floor and ceiling. So for those, repeat the above, except use \varepsilon = 1 and T = 20 deg C + 273.15 K = 293.15 K.
For the rest, you're going to have to get a little creative in the interpretation. The human body has nooks and crannies and overlap. For example, some of the energy radiating from your the sides of your torso might hit the sides of your arms and vice versa. So using your body's "true" surface shape isn't particularly helpful.
So what we're going to do is approximate your body as a cylinder or sphere or some shape that doesn't overlap back on itself. In other words, if a bit a radiation leaves your body it will continue away from you until it strikes a wall/floor/ceiling, without hitting some other part of your own body. I'll assume a cylinder as we continue.
With these assumptions, the radient emittance leaving the walls is excactly the same thing as the irradiance at the surface of your body. Note that irradiance is the radiant flux received by a surface per unit area (example units are \mathrm{W/m^2}). Note that irradiance is also a flux density and has the same units as radiant emittance. So since you're completely surrounded by the walls and ceiling, your irradiance (the energy flux density striking your body) is the same as the emittance of the walls. What I'm saying is that the emittance of the walls is the answer to the second part of the problem, your average thermal irradiance at your surface.
The last tricky bit is that some of the energy striking your body gets reflected. As it happens, the fraction of radiation that your body absorbs (isn't reflected) is
your emissivity.
So your final, net radiant heat loss becomes,
\mathrm{Net \ radiant \ heat \ loss} = \left( M_{e\_you} - \varepsilon M_{e\_walls} \right) A,
where \varepsilon is
your emissivity (not the walls' emissivity), here 0.97, and A is the surface area of the cylinder approximating your body. The units of the net radiant heat loss are units of power, such as Watts. (It is possible to simplify the above equation such that it doesn't use intermediate variables, but I'll leave that up to the reader.)
As for the dimensions of the cylinder, I don't know, but something like r = 0.2 m, and h = 1.6 m sounds like a fair starting point. Whatever dimensions you choose, be sure to document your approach, and I'm sure your instructor will understand.